Percent Yield Worksheet 2 - Free Printable

Here are the step-by-step solutions to the problems on your worksheet.

Problem 1: Balancing and Reaction Type

Equation: $\text{Mg} + \text{HNO}_3 \rightarrow \text{Mg(NO}_3)_2 + \text{H}_2$

* Balancing:
1. Look at Nitrate ($\text{NO}_3$). There is 1 on the left and 2 on the right. Put a 2 in front of $\text{HNO}_3$.
2. Now check Hydrogen (H). There are 2 on the left (from $2\text{HNO}_3$) and 2 on the right ($\text{H}_2$). It is balanced.
3. Check Magnesium (Mg). There is 1 on the left and 1 on the right. It is balanced.
* Coefficients: 1, 2, 1, 1

* Type of Reaction:
This is a reaction where an element (Mg) replaces another element (H) in a compound. This is called a Single Replacement (or Single Displacement) reaction.

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Problem 2: Theoretical Yield Calculation

Question: If you start with 40 grams of magnesium (Mg) and excess nitric acid, how many grams of hydrogen gas ($\text{H}_2$) will be produced?

1. Find Molar Masses:
* Mg = $24.31 \text{ g/mol}$
* $\text{H}_2$ = $1.008 \times 2 = 2.016 \text{ g/mol}$

2. Convert Grams of Mg to Moles of Mg:
$$40 \text{ g Mg} \div 24.31 \text{ g/mol} = 1.645 \text{ moles Mg}$$

3. Use the Mole Ratio:
From the balanced equation ($1\text{Mg} \rightarrow 1\text{H}_2$), the ratio is 1:1.
So, we produce 1.645 moles of $\text{H}_2$.

4. Convert Moles of $\text{H}_2$ to Grams:
$$1.645 \text{ moles} \times 2.016 \text{ g/mol} = 3.317 \text{ grams}$$

Answer: You will produce approximately 3.32 grams of hydrogen gas.

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Problem 3: Percent Yield

Question: If 1.7 grams of hydrogen is actually produced, what was the percent yield?

* Formula: $\text{Percent Yield} = (\frac{\text{Actual Yield}}{\text{Theoretical Yield}}) \times 100$
* Actual Yield: 1.7 g (given in the problem)
* Theoretical Yield: 3.32 g (calculated in Problem 2)

$$ \text{Calculation: } (\frac{1.7}{3.32}) \times 100 = 51.2\% $$

Answer: The percent yield is 51.2%.

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Problem 4: Balancing and Reaction Type

Equation: $\text{NaHCO}_3 \rightarrow \text{NaOH} + \text{CO}_2$

* Balancing:
1. Count atoms on the left: Na=1, H=1, C=1, O=3.
2. Count atoms on the right: Na=1, H=1, C=1, O=3 ($1 \text{ in NaOH} + 2 \text{ in CO}_2$).
3. Everything is already equal!
* Coefficients: 1, 1, 1

* Type of Reaction:
One single compound breaks down into two simpler substances. This is a Decomposition reaction.

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Problem 5: Stoichiometry Calculation

Question: If 25 grams of carbon dioxide ($\text{CO}_2$) gas is produced, how many grams of sodium hydroxide ($\text{NaOH}$) should be produced?

1. Find Molar Masses:
* $\text{CO}_2$ = $12.01 + (16.00 \times 2) = 44.01 \text{ g/mol}$
* $\text{NaOH}$ = $22.99 + 16.00 + 1.008 = 40.00 \text{ g/mol}$

2. Convert Grams of $\text{CO}_2$ to Moles:
$$25 \text{ g} \div 44.01 \text{ g/mol} = 0.568 \text{ moles CO}_2$$

3. Use the Mole Ratio:
From the balanced equation ($1\text{NaHCO}_3 \rightarrow 1\text{NaOH} + 1\text{CO}_2$), the ratio of $\text{CO}_2$ to $\text{NaOH}$ is 1:1.
So, we produce 0.568 moles of $\text{NaOH}$.

4. Convert Moles of $\text{NaOH}$ to Grams:
$$0.568 \text{ moles} \times 40.00 \text{ g/mol} = 22.72 \text{ grams}$$

Answer: You should produce 22.7 grams of sodium hydroxide.

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Problem 6: Percent Yield

Question: If 50 grams of sodium hydroxide are actually produced, what was the percent yield?

* Formula: $\text{Percent Yield} = (\frac{\text{Actual Yield}}{\text{Theoretical Yield}}) \times 100$
* Actual Yield: 50 g (given in the problem)
* Theoretical Yield: 22.7 g (calculated in Problem 5)

$$ \text{Calculation: } (\frac{50}{22.7}) \times 100 = 220.2\% $$

*(Note: In real life, getting over 100% usually means the product wasn't dried properly and still had water weight, but mathematically, this is the correct answer based on the numbers provided.)*

Answer: The percent yield is 220%.